Two Sum

Problem Link

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

You can return the answer in any order.

Example 1:

• Input: nums = [2,7,11,15], target = 9
• Output: [0,1]
• Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].

Example 2:

• Input: nums = [3,2,4], target = 6
• Output: [1,2]

Example 3:

• Input: nums = [3,3], target = 6
• Output: [0,1]

Constraints:

• 2 <= nums.length <= 10⁴
• -10⁹ <= nums[i] <= 10⁹
• -10⁹ <= target <= 10⁹
• Only one valid answer exists.

Follow-up: Can you come up with an algorithm that is less than O(n2) time complexity?

#include <vector>
#include <unordered_map>
using namespace std;

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> numMap; // Hash table to store number and its index

        for(int i = 0; i < nums.size(); i++) {
            int complement = target - nums[i]; // The number that needs to pair with the current number

            if(numMap.find(complement) != numMap.end()) {
                // Add the current number to the hash table
                return { numMap[complement], i };
            }
            numMap[nums[i]] = i;
        }

        // Assuming there is an answer, this part will not be executed
        return {};
    }
};